The Chinese Remainder Theorem

Transcription

1 Sacred Heart University March 29, 2018

2 Divisibility Divisibility We say a divides b, denoted as a b, if there exists k Z such that ak = b. Example: Consider 2 6. Then k = 3 since (2)(3) = 6. Property - Addition Let a, b, and c Z. If a b and a c then a (b+c). Proof. ak = b am = c Then a(k + m) = b + c. This implies a (b + c).

3 Properties of Divisibility Property - Subtraction Let a, b, and c Z. If a b and a c then a (b-c). Property - Multiplication Let a, b, and c Z. If a b then a bc. Property - Transitive Let a, b and c Z. If a b and b c, then a c.

4 Division Algorithm and GCD Division Algorithm Let m and n N. Then there exist unique integers q, r such that m = qn + r, where 0 r n - 1. Great Common Divisior The Greatest Common Divisor of two integers a and b, not both 0, is the largest, positive integer that divides a and b. It is denoted as (a, b).

5 Example of the Euclidean Algorithm Theorem leading to the Euclidean Algorithm Let a, n, b, and r Z where a and b are not both 0. If a = nb + r, then (a,b) = (b, r). gcd(124, 144). 144 = (124 1) = (20 6) = (4 5) + 0. Then the gcd(124, 144) = 4. Retracing our steps, we express the gcd in a linear combination of its terms 124 and 144: 4 = (20 6) 4 = (( ) 6) 4 = (6 144) + (6 124) 4 = (- 6) (7) 124

6 A Linear Diophantine Equation A linear Diophantine equation ax + by = c A linear equation of this form ax + by = c of two variables is a linear Diophantine equation. GCD is a linear combination Let a and b Z not both 0. Then there exist x and y Z such that ax+by= (a,b). Theorem Let a, b and c Z. If a bc and (a,b) = 1, then a c. Proof. Since (a, b) = 1, then ax + by = 1. Then, cax + cby = c Since a cax and a cby, then a (cax + cby). Therefore a c.

7 Condition for solutions for a linear Diophantine equation Let a, b, and c Z with a and b not both 0, then there exist x, y Z that satisfy the equation ax + by = c if and only if (a,b) c. Proof. Suppose ax + by = c and (a,b) = d. Then d a and d b. This implies that d c Conversely, assume d c. Then kd = c Then let d = (a, b), This implies ax + by = d. kd = k(ax + by ) a(kx ) + b(ky ) = c. Let x = kx and y = ky. Therefore this satisfies the equation ax + by = c.

8 Example of a Linear Diophantine Equation Consider the linear Diophantine equation of two variables: 3x + 6y = 3. We start with any solution, say (1,0).

9 Linear Congruences Congruence Modulo n We say a is congruent to b (mod n), if n (a - b). It is denoted as a b (mod n). A linear congruence modulo n has a solution if its Diophantine equation has a solution. Theorem Let a, b, and n be integers with n > 0. The congruence ax b (mod n) has a solution if and only if there exist integers x and y such that ax + ny = b.

10 Proof. Assume ax b (mod n) has a solution. Then n (ax - b) n -(ax - b). This implies ny = b - ax. Rewriting we have ax + ny = b. Conversely, assume ax 0 + ny 0 = b. Then ax 0 - b = n(-y 0 ). This implies ax 0 b(mod n) has a solution.

11 Solution to Linear Congruences Condition for a linear congruence to have a solution Let a, b, and n be integers with n > 0. The equation ax b (mod n) has a solution if and only if (a, n) b. Proof. Suppose ax b (mod n) has a solution x 0. ax + by = b (a, n) b This implies ax b (mod n) has a solution if and only if (a, n) b.

12 Solutions to a System of Linear Congruences Condition for a system of linear congruences to have a solution Let a, b, m and n be integers with m > 0 and n > 0. Then the system x a (mod n), x b (mod m) has a solution if and only if (n, m) (a - b). Proof. Suppose there is a solution to the system x a (mod n) and x b (mod m). Let d = (n, m). We have d n and d m We also know n (x - a) and m (x - b). This implies d (x - a) and d (x - b). Subtracting we have d ((x - b) - (x - a)) Therefore d (a - b).

13 Solutions to a System of Linear Congruences Proof. Suppose (n, m) (a - b). Then there exists x 0 Z such that mx 0 (a - b)(mod n). This implies mx 0 + b a (mod n). Also mx 0 + b b (mod m), thus x = mx 0 + b is a solution to the system.

14 A system of two linear congruences relatively prime Let a, b, m and n be integers with m > 0, n > 0 and (n, m) = 1. Then the system x a (mod n), x b (mod m) has a unique solution modulo mn. Proof. The system has a solution say x 0 because 1 (a - b). Let x a (mod n) and x b (mod m) be another solution to the system. Then x 0 - x 0 (mod n) And x 0 - x 0 (mod m). This implies n (x 0 - x ) and m (x 0 - x ) Since (m, n) = 1, then mn (x 0 - x ) Thus x 0 x (mod mn). This implies x 0 is the unique solution.

15 Generalization Suppose n 1, n 2... n L are positive integers that are pairwise relatively prime, that is (n i, n j ) = 1 for i j, 1 i, j L. Then the system of L congruences x a 1 (mod n 1 ), x a 2 (mod n 2 )... x a L (mod n L ) has a unique solution modulo the product of n 1 n 2 n 3...n L.

16 Proof. Induction step: Suppose the system of L - 1 congruences x a 1 (mod n 1 ), x a 2 (mod n 2 )... x a L 1 (mod n L 1 ) has a unique solution modulo n 1...n L 1 say x 0. Then we have x x 0 (mod n 1 n 2...n L 1 ). We need to prove the result is true for L congruences: x a 1 (mod n 1 ), x a 2 (mod n 2 )... x a L 1 (mod n L 1 ), x a L (mod n L ) where (n i, n j ) = 1 for all i, j such that 1 i, i L. Since n 1, n 2,..., n L 1,n L are pairwise relatively prime, it follows that the product n 1 n 2...n L 1 is relatively prime to n L. Since x x 0 (mod n 1 n 2...n L 1 ) and x a L (mod n L ) which is a pair of relatively prime congruences, it follows that this system has a unique solution mod n 1 n 2...n L 1 n L, say x. Then x is the unique solution modulo to the system x a 1 (mod n 1 ), x a 2 (mod n 2 )... x a L (mod n L ).

17 Example of Easter Eggs! Let s make an Easter egg example. Sally, Billy, and Ann are children that have collected eggs left behind from the Easter Bunny. The system of congruences is as followed: x 5 (mod 13), x 7 (mod 12), and x 0 (mod 11). What is the least amount of eggs they ve collected?

18 Example of Easter Eggs! Let s make an Easter egg example. Sally, Billy, and Ann are children that have collected eggs left behind from the Easter Bunny. The system of congruences is as followed: x 5 (mod 13), x 7 (mod 12), and x 0 (mod 11). What is the least amount of eggs they ve collected? Solution 13k = x k (mod 12). k 2 (mod 12) or k = 12m (12m + 2) = x - 5 or x = 156m m (mod 11) m 1 (mod 11) or m = j x = 156(1 + 11j) + 31 or x = 1716j Our least solution is 187 (mod 1716).

19 Further Research and Questioning considers a system of only pairwise relatively prime modulo congruences, but what about systems of linear congruences with moduli with gcd > 1? Is there a systematic method or theorem that describes solutions of congruences not pairwise, relatively prime modulo?

20 References Marshall D., Odell E. and Starbird M., 2007: Number Theory Through Inquiry. The Mathematical Association of America, 151.

21 The End